After going through multiple youtube videos and some articles I'm still having trouble imagining voltage.
If we measure across a battery's terminals we will see electrical potential. This is caused by the collection of negative particles on the negative terminal and positive particles on the positive terminal. Since these are being "kept apart", there is some work being done to hold them where they are, and therefore potential energy. The difference in potential energy between the two terminals is our voltage.
When we connect a battery to a simple circuit consisting of a lamp (for example), we can measure the two points on the conductor using a voltmeter and see 0V. Since the electrons are being "pulled" towards the positive terminal (in the same sense that gravity pulls a dropped object downward), there is no work being done to prevent the from traveling to the positive terminal, therefore the difference in the potential energies between the two points is 0.
Now when we measure across the lamp in the circuit we will once again see a voltage. I think this is where I get confused. The lamp isnt part of the conductor, and therefore there is voltage – this is where my book leaves off.
Using my reasoning from above – is this because on the positive end of the lamp, it's connected to the positive terminal and has a higher positive charge than the rest of the conductor? This would explain why there is a measurable potential difference. If so, what happens inside of the lamp with the potential energy? Some of it must be turned to heat?
Best Answer
You have a fundamental misunderstanding about voltage. A voltage difference isn't caused by an imbalance in the amount of charge, it's caused by a difference in the amount of energy available from the charge.
As charge passes (i.e. current flows) through an incandescent lamp it loses electrical energy, which is converted to heat and light. The density of electrons is the same at both ends of the filament, but the charge entering the filament has more energy than the charge leaving the filament.